Let $k<d$ be positive integers. For which values of $1<k<d$ (if any), does there exist an integer $1<r<d$, such that $\binom{d}{r}^2\cdot d=\binom{d}{k-1}\binom{d}{k}$?
$r=d$, $k=1$ work. Trying to plug $k=r$ or $k=r+1$ does not provide any other solutions. I guess that there are no other solutions, but I don't know how to prove this.
The motivation for this question comes from counting degrees of freedom, in an attempt to find a variational interpretation for an equation.